Find the gradient of the line passing through the points P(1, 1) and Q(2, 5).
3
2
5
4
Correct answer is D
Let (x1, y1) = (1, 1) and (x2, y2)= (2, 5)
then gradient m of \(\bar{PQ}\) is
m = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{5 - 1}{2 - 1}\)
= \(\frac{4}{1}\)
= 4
1
0
√3
√2
Correct answer is D
Let D denote the distance between (\(\frac{1}{2}\), -\(\frac{1}{2}\)) then using
D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
= \(\sqrt{(-{\frac{1}{2} - \frac{1}{2}})^2 + (-{\frac{1}{2} - \frac{1}{2}})^2}\)
= \(\sqrt{(-1)^2 + (-1)^2}\)
= \(\sqrt{1 + 1}\)
= √2
749\(\pi\)m2
700\(\pi\)m2
350\(\pi\)m2
98\(\pi\)m2
Correct answer is A
Total surface area of the cylindrical pipe = area of circular base + curved surface area
= \(\pi\)r\(^2\) + 2\(\pi\)rh
= \(\pi\) x 7\(^2\) + 2\(\pi\) x 7 x 50
= 49\(\pi\) + 700\(\pi\)
= 749\(\pi\)m\(^2\)
112o
102o
82o
52o
Correct answer is D
(x + 15)° + (2x - 45)° + (x + 10)° = (2n - 4)90°
when n = 4
x + 15° + 2x - 45° + x - 30° + x + 10° = (2 x 4 - 4) 90°
5x - 50° = (8 - 4)90°
5x - 50° = 4 x 90° = 360°
5x = 360° + 50°
5x = 410°
x = \(\frac{410^o}{5}\)
= 82°
Hence, the value of the least interior angle is (x - 30°)
= (82 - 30)°
= 52°
If P = \(\begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix}\) , what is P\(^-1\)
\(\begin{pmatrix} -{\frac{1}{5}} & -{\frac{3}{5}} \\ -{\frac{1}{5}} & -{\frac{2}{5}} \end{pmatrix}\)
\(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \\ {\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\)
\(\begin{pmatrix} -{\frac{1}{5}} & {\frac{3}{5}} \\ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\)
\(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \\ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\)
Correct answer is D
P = \(\begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix}\)
|P| = 2 - 1 x -3 = 5
P-1 = \(\frac{1}{5}\)\(\begin{pmatrix} 1 & 3 \\ -1 & 2 \end{pmatrix}\)
= \(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \\ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\)